Let's say I have the following data frame of series.
library(zoo)
n=10
date = as.Date(1:n);date
y = rnorm(10);y
dat = data.frame(date,y)
dat
date y
1 1970-01-02 -0.02052313
2 1970-01-03 0.28255304
3 1970-01-04 -0.10718621
4 1970-01-05 -1.19299366
5 1970-01-06 1.17072468
6 1970-01-07 0.55849119
7 1970-01-08 0.30474050
8 1970-01-09 -0.30777180
9 1970-01-10 -0.01874367
10 1970-01-11 -0.74233556
Calculating the rolling standard deviation with width (rolling window) of 3 days,i do:
a = zoo::rollapply(dat$y,3,sd)
With results:
2 3 4 5 6
1.7924102 0.4189522 0.4599979 0.4164408 0.3786601
7 8 9
0.9481849 0.7048868 0.2494578
And finding the maximum standard deviation
max(a)
1.79241
Now I want to find out in which time 3-day interval this maximum refers to.How can I do that? Imagine that my series is 20 years, so I want to find the maximum standard deviation of this 20 years and extract the specific 3 day time interval.
Updated in response to OPs clarification of working days to include a date vector which misses out weekends, in a nominal manner (I've not checked if these dates are real weekends, the purpose is to miss out dates for weekends).
set.seed(123)
n = 10
date = as.Date(c(1, 4:8, 11:14), origin = "1970-01-01")
y = rnorm(10)
dat = data.frame(date ,y)
roll_win <- 3
dat$a = c(rep(NA_real_, roll_win - 1), zoo::rollapply(dat$y, roll_win ,sd))
dat <- subset(dat, !is.na(a))
dat_max <- dat[dat$a == max(dat$a, na.rm = TRUE), ]
dat_max$date_start <- dat$date[which(dat$a == max(dat$a, na.rm = TRUE)) - (roll_win - 1)]
dat_max
#> date y a date_start
#> 8 1970-01-13 -1.265061 1.496275 1970-01-09
Created on 2022-02-22 by the reprex package (v2.0.1)